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Let $g$ a Lie algebra and $V$ a finite-dimensional irreducible $g$-module, then each generalized eigenspace of $V$ is actually an eigenspace? If not, what is a condition to guarantee this fact?

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Please do not cross-post: mathoverflow.net/questions/55871/… –  Qiaochu Yuan Feb 18 '11 at 16:35
    
Each generalized eigenspace of what? –  Mariano Suárez-Alvarez Feb 18 '11 at 16:43
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I am sorry but that did not help: what is a "generalized eigenspace of $V$"? –  Mariano Suárez-Alvarez Feb 18 '11 at 16:52
    
Let $\rho: g \mapsto gl(V)$ such representation. The generalized eigenspace of $V$ via $\rho$ is $$V_i=\left\{v\in V \mid \forall x\in g, \exists n\in \mathbb N \text{ such that } (\rho(X) - \lambda_i(X))^nv = 0\right\}.$$ –  Binai Feb 18 '11 at 17:05
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@user2764: add that to the body of the question, so that everybody can see it without having to read through all the comments. Presumably, you fixed a linear form $\lambda:g\to\mathbb C$ and are describing the "generalized eigenspace corresponding to $\lambda$"? –  Mariano Suárez-Alvarez Feb 18 '11 at 18:39
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I am not sure I get exactly what you mean, but the answer is probably no.

If $g$ is one-dimensional and we fix a non-zero element $X_0\in g$, then providing a $g$-module structure to a vector space $V$ is the same thing as picking an endomorphism $T\in\gl(V)$, corresponding to the action of $X_0$ on the module (i.e., what you call $\rho(X_0)$ is a comment above).

If you pick a linear map $T:V\to V$ which is not semisimple (that is, whose Jordan canonical form is not diagonal), then your «generalized eigenspaces of $V$» are surely the same thing as the generalized eigenspaces of $T$, and these are certainly not actual eigenspaces, because of the choice of $T$.

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